\(\displaystyle \int \limits_1^2 \dfrac{1}{x^2}   dx = \)

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Rewriting the integrand,

$$ \int \limits_1^2 \dfrac{1}{x^2}   dx = \int \limits_1^2 x^{-2}   dx =$$ $$ = \Big[ -x^{-1} \Big]_1^2 = \left[ -\frac{1}{x} \right]_1^2 $$ $$ = -\left( \frac{1}{2} - \frac{1}{1} \right) $$ $$ = \frac{1}{2} $$